Consider the polar curve $r=\cos(\theta)$. What is the slope of the tangent line to the curve $r$ when $\theta = \dfrac{\pi}{4}$ ? Give an exact expression. $\text{slope }=$
Explanation: The slope of the tangent line at a point is equal to $\dfrac{dy}{dx}$ at that point. In the case of polar curves, we can use the relationship: $\dfrac{dy}{dx}=\dfrac{\left( \dfrac{dy}{d\theta} \right)}{\left( \dfrac{dx}{d\theta} \right)}$ For a polar curve, $x={r}\cos(\theta)$ and $y={r}\sin(\theta)$. Therefore, in our problem we have: $\begin{aligned} x&={\cos(\theta)}\cos(\theta) \\\\ &=\cos^2(\theta) \\\\ y&={\cos(\theta)} \sin(\theta) \\\\ &=0.5\sin(2\theta) \end{aligned}$ Let's find $\dfrac{dy}{dx}$. $\begin{aligned} \dfrac{dy}{dx}&=\dfrac{\left( \dfrac{dy}{d\theta} \right)}{\left( \dfrac{dx}{d\theta} \right)} \\\\ &=\dfrac{\cos\left(2\theta\right)}{-\sin\left(2\theta\right)} \\\\ &=-\dfrac{\cos\left(2\theta\right)}{\sin\left(2\theta\right)} \end{aligned}$ Finally, we evaluate $\dfrac{dy}{dx}$ at ${\theta = \dfrac{\pi}{4}}$. $\begin{aligned}\left. \dfrac{dy}{dx} \right| _{\theta =\tfrac{\pi }{4}}&=-\dfrac{\cos\left(2\left({\dfrac{\pi}{4}}\right)\right)}{\sin\left(2\left({\dfrac{\pi}{4}}\right)\right)} \\\\ &=-\dfrac{\cos\left(\dfrac{\pi}{2}\right)}{\sin\left(\dfrac{\pi}{2}\right)} \\\\ &=-\dfrac{0}{1} \\\\ &=0 \end{aligned}$ The slope of the tangent line to the curve $r$ when $\theta=\dfrac{\pi}{4}$ equals $0$. The graph of the tangent is shown. ${0.5}$ ${1}$ ${1.5}$ ${0}$ ${\frac{1}{8}\pi}$ ${\frac{1}{4}\pi}$ ${\frac{3}{8}\pi}$ ${\frac{1}{2}\pi}$ ${\frac{5}{8}\pi}$ ${\frac{3}{4}\pi}$ ${\frac{7}{8}\pi}$ ${\pi}$ ${\frac{9}{8}\pi}$ ${\frac{5}{4}\pi}$ ${\frac{11}{8}\pi}$ ${\frac{3}{2}\pi}$ ${\frac{13}{8}\pi}$ ${\frac{7}{4}\pi}$ ${\frac{15}{8}\pi}$